A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #42 #. What is the area of the triangle's incircle?

The incircle is tangent to all the sides. The incenter, its center, is the meet of the angle bisectors.

When I have trouble getting started I pin my triangle to the Cartesian plane. #C(0,0), A(a,0), B(b,c)#.

Let's express our facts;

#A=\pi/8#

#B=pi/4#

#C = \pi-pi/8-pi/4={5pi}/8#

# tan A = c/a #

# tan C = c/b #

Let's call the area #mathcal{A=42}#.

# mathcal{A} = 1/2 a c #

That's three equations in three unknowns. We mostly care about #a#.

# 2 mathcal{A} = ac = a (a tan A)=a^2 tan A #

# a = \sqrt{ {2 mathcal{A} }/ tan A} #

Tangents are slopes. The perpendicular bisector of C has equation

#y = x tan (C/2)#

The perpendicular bisector of A has equation

#y = -(x-a) tan (A/2)#

Those meet at the incenter, when

#x tan(C/2) = -x tan(A/2) + a tan (A/2)#

#x = {a tan(A/2)}/{tan(A/2) + tan(C/2)}#

#y = {a tan(A/2) tan(C/2)}/{tan(A/2) + tan(C/2)}#

The incenter is #(x,y)# and the incircle is tangent to AC, the x axis, so the radius is #y# and the incircle's area is #pi y^2#.

Let's put everything together.

#text{incircle_area} = {2 pi mathcal{A} tan^2(A/2) tan^2(C/2)}/{ tan A(tan(A/2) + tan(C/2))^2}#

There's probably some simplification to be done here, but let's plug in the numbers to see if I've gone off the deep end. I'd guess from my rough sketch an area a little less than half, around 20.

Feeding it to Alpha gets an approxmate incircle area of #19.6399#, closer to my guess than I would have guessed. Hard to believe I didn't make a mistake getting this far, but let's proceed.

That's motivation to work out the exact answer but I'm getting length warnings, so let's stop here.